cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-22 of 22 results.

A308734 Number of ordered ways to write n as (2^a*3^b)^2 + (2^c*5^d)^2 + x^2 + y^2, where a,b,c,d,x,y are nonnegative integers with x <= y.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 3, 1, 3, 5, 2, 3, 4, 4, 5, 1, 4, 8, 4, 4, 8, 8, 4, 3, 8, 7, 7, 6, 5, 13, 6, 1, 10, 11, 7, 7, 10, 9, 9, 5, 7, 18, 7, 5, 14, 11, 6, 3, 10, 11, 9, 8, 7, 15, 9, 4, 14, 12, 5, 10, 9, 10, 11, 1, 11, 19, 10, 6, 17, 21, 6, 8, 14, 12, 13, 7, 14, 21, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Jun 21 2019

Keywords

Comments

Four-square Conjecture: a(n) > 0 for all n > 1.
This is much stronger than Lagrange's four-square theorem. We have verified a(n) > 0 for all n = 2..10^9.
Note that 16265031 cannot be written as (2^a*3^b)^2 + (2^c*3^d)^2 + x^2 + y^2 with a,b,c,d,x,y nonnegative integers.
a(n) > 0 for 1 < n <= 10^10. - Giovanni Resta, Jun 28 2019
I promise to offer 2500 US dollars as the prize for the first correct proof of the Four-square Conjecture. - Zhi-Wei Sun, Jul 09 2019
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 1.6*10^11. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(2^(2k+1)) = 1 with 2^(2k+1) = (2^k*3^0)^2 + (2^k*5^0)^2 + 0^2 + 0^2.
a(2^(2k+2)) = 1 with 2^(2k+2) = (2^k*3^0)^2 + (2^k*5^0)^2 + (2^k)^2 + (2^k)^2.
a(3) = 1 with 3 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
a(5) = 2 with 5 = (2^0*3^0)^2 + (2^1*5^0)^2 + 0^2 + 0^2 = (2^1*3^0)^2 + (2^0*5^0)^2 + 0^2 + 0^2.
a(11) = 2 with 11 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 3^2 = (2^0*3^1)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[n-4^a*9^b-4^c*25^d-x^2],r=r+1],{a,0,Log[4,n]},{b,0,Ceiling[Log[9,n/4^a]]-1},
    {c,0,Log[4,n-4^a*9^b]},{d,0,Log[25,(n-4^a*9^b)/4^c]},{x,0,Sqrt[(n-4^a*9^b-4^c*25^d)/2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A305232 Number of ordered ways to write 2*n+1 as p + binomial(2k,k) + 2*binomial(2m,m), where p is an odd prime, and k and m are nonnegative integers.

Original entry on oeis.org

0, 0, 1, 2, 3, 3, 3, 4, 3, 5, 4, 5, 6, 5, 4, 4, 6, 6, 4, 5, 4, 6, 6, 6, 7, 6, 6, 4, 5, 6, 6, 8, 5, 5, 6, 5, 7, 9, 8, 5, 8, 9, 6, 9, 7, 8, 6, 6, 4, 7, 8, 7, 7, 4, 8, 10, 9, 7, 8, 9, 5, 7, 6, 5, 7, 7, 7, 3, 6, 7, 7, 9, 6, 9, 6, 9, 9, 7, 7, 8, 9, 6, 5, 8, 10, 10, 6, 8, 7, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 27 2018

Keywords

Comments

The first value of n > 2 with a(n) = 0 is 15212443837. Neither 2*15212443837 + 1 = 30424887675 nor 2*15657981007 + 1 = 31315962015 can be written as the sum of a prime, a central binomial coefficient and twice a central binomial coefficient.

Examples

			a(3) = 1 since 2*3 + 1 = 7 = 3 + binomial(2*1,1) + 2*binomial(2*0,0) with 3 an odd prime.
a(368233372) = 1 since 2*368233372 + 1 = 736466745 = 735761311 + binomial(2*11,11) + 2*binomial(2*0,0) with 735761311 an odd prime.
a(5274658504) = 1 since 2*5274658504 + 1 = 10549317009 = 10549316083 + binomial(2*6,6) + 2*binomial(2*0,0) with 10549316083 an odd prime.
a(8722422187) = 1 since 2*8722422187 + 1 = 17444844375 = 17444844367 + binomial(2*2,2) + 2*binomial(2*0,0) with 17444844367 an odd prime.
a(10296844792) = 1 since 2*10296844792 + 1 = 20593689585 = 20593688659 + binomial(2*6,6) + 2*binomial(2*0,0) with 20593688659 an odd prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;k=0;Label[aa];k=k+1;If[Binomial[2k,k]>=2n+1`,Goto[cc]];m=0;Label[bb];If[2*Binomial[2m,m]>=2n+1-Binomial[2k,k],Goto[aa]]; If[PrimeQ[2n+1-Binomial[2k,k]-2*Binomial[2m,m]],r=r+1];m=m+1;Goto[bb];Label[cc];tab=Append[tab,r],{n,1,90}];Print[tab]
Previous Showing 21-22 of 22 results.